我对 Ajax 有点陌生，我一直在尝试找出我做错了什么。我从数据库中提取结果并放入 xml。当我通过 xml 循环时，我正在尝试执行一个 php 文件，同时将 xml 结果中的 ID 号发送给它，然后从 php 文件中返回“echo”。我不确定我是完全关闭还是只是缺少一部分，但结果返回“未定义”。
这是我试图回显并显示出来的 php 文件。
echo rating_bar($id);
function rating_bar($id) {
//other code, but $static_rater is what gets echoed
$static_rater = "";
$static_rater .= '<div id="ratingblock" class="ratingblock">';
$static_rater .= '<div id="unit_long'.$id.'">';
$static_rater .= '<ul id="unit_ul'.$id.'" class="unit-rating" style="width:'.$rating_unitwidth*$units.'px;">';
$static_rater .= '<li class="current-rating" style="width:'.$rating_width.'px;"></li>';
$static_rater .= '</ul>';
$static_rater .= '<p class="static">Rating: <strong> '.$rating1.'</strong>/'.$units.' ('.$count.' '.$tense.' cast)</p>';
$static_rater .= '</div>';
$static_rater .= '</div>';
//return join("\n", $static_rater);
echo $static_rater;exit;
}
这是我试图返回结果的 .js 代码。
downloadUrl("phpsqlajax_genxml.php", function(data) {
var xml = data.responseXML;
var bounds = new google.maps.LatLngBounds();
var markers = xml.documentElement.getElementsByTagName("marker");
// alert("downloadUrl callback, length="+markers.length);
for (var i = 0; i < markers.length; i++) {
var id = markers[i].getAttribute("id");
if (!id) id = "id "+i;
var name = markers[i].getAttribute("name");
if (!name) name = "name "+i;
var address = markers[i].getAttribute("address");
if (!address) address = "address";
var citystate = markers[i].getAttribute("citystate");
if (!citystate) citystate = "city, ST";
var phone = markers[i].getAttribute("phone");
if (!phone) phone = "phone number";
var type = markers[i].getAttribute("type");
var point = new google.maps.LatLng(
parseFloat(markers[i].getAttribute("lat")),
parseFloat(markers[i].getAttribute("lng")));
bounds.extend(point);
var html = "<b>" + name + "</b> <br/>" + address + "<br/>" + citystate + "<br/>" + phone; //html inside InfoWindow
var url = "starrating/_drawrating.php?id=" + id + "";
//var contentString = ajaxLoad(url, parseResults, true);
//var contentString = downloadUrl(url, "POST", "text=" + text, completed);
var contentString = AJAX('starrating/_drawrating.php','id='+id,
function(data) {
var htm = $("#ratingblock").html(data);
alert(htm);
}
);
var description = "<br><br>description" + id + " <br><b>" + name + "</b> <br/>" + address + "<br/>" + citystate + "<br/>" + phone; //html inside InfoWindow
var icon = customIcons[type] || {};
var marker = new google.maps.Marker({
map: map,
position: point,
icon: icon.icon,
shadow: icon.shadow,
animation: google.maps.Animation.DROP
});
bindInfoWindow(marker, map, infoBubble, html, description, contentString);
}
});
function AJAX(url, data, callback)
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
callback(xmlhttp.responseText);
}
}
xmlhttp.open("POST",url,true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send(data);
}
编辑：好的，所以我一直在玩这个并更新了我上面的代码。现在，当我使用 firebug 运行它时，我可以看到帖子并且响应是这样的：
<div id="ratingblock" class="ratingblock"><div id="unit_long10"><ul id="unit_ul10" class="unit-rating" style="width:150px;"><li class="current-rating" style="width:0px;"></li></ul><p class="static">Rating: <strong> 0.0</strong>/5 (0 votes cast)</p></div></div>
但警报只显示 [object Object]，而 infowindow 显示 [object Object]。所以我知道它的调用和返回数据，我搜索并尝试了我能想到的一切，以使上述部分正确显示在信息窗口中。有什么想法吗？
编辑#2
我在下面尝试一种新方法。
var contentString = $.ajax({
type:"POST",
url: "starrating/_drawrating.php",
dataType: "html",
data:"id="+id,
success: function(data){
var $response=$(data);
$response.find('ratingblock').html();
console.log($response);
}
});
控制台返回“Object[div#ratingblock.ratingblock]”，但结果仍然显示 [object Object]。任何想法我缺少什么？